Question

Calculate the value of mean free path $\left(\lambda \right)$ for oxygen molecules at temperature $27°C$ and pressure $1.01\times {10}^{5}Pa$. Assume the molecular diameter $0.3nm$ and the gas is ideal. $\left(1.38\times {10}^{-23}j{K}^{-1}\right)$.

• A. $102nm$
• B. $32nm$
• C. $58nm$
• D. $86nm$

Answer 1

The correct option is A

$102nm$

Step 1: Given data

Pressure, $P=1.01\times {10}^{5}Pa$

Diameter, $d=0.3nm=0.3\times {10}^{-9}m$

Temperature, $$\begin{gathered} T=27°C=\left(27+273\right)K \\ T=300K \end{gathered}$$

Boltzmann constant, $k=1.38\times {10}^{-23}j{K}^{-1}$

Step 2: Finding the value of mean free path, $\lambda$

By using the formula of the mean free path,

$\lambda =\frac{kT}{\sqrt{2}{\mathrm{\pi d}}^2\mathrm{P}}$ [ Where, $$\begin{gathered} d=0.3\times {10}^{-9}m \\ T=300K \end{gathered}$$]

$\lambda =\frac{1.38\times {10}^{-23}\times 300}{\sqrt{2}\times 3.14\times 0.09\times {10}^{-18}\times 1.01\times {10}^5}$

$\lambda =102nm$

Hence, the correct option is $\left(A\right)$.